Question

A projectile is fired vertically upwards from the surface of the earth with a velocity $K{v}_e$, where $v_0$ is the escape velocity and $k<1$. If $R$ is the radius of the earth, the maximum height to which it will rise measured from the centre of the earth, will be (neglect air resistance).

• A. $\frac{1-k^2}{R}$
• B. $\frac{R}{1-k^2}$
• C. $R(1-k{)}^2$
• D. $\frac{R}{1+k^2}$

Answer 1

Answer: (B)

$\frac{R}{1-k^2}$

If a body is projected from the surface of the earth with a velocity $v$ and reaches a height $h$, then according to law of conservation of energy.
$\frac{1}{2}m{v}^2=\frac{mgh}{1+h/R}$
Here $v=k{v}_e=k\sqrt{2gR}$

Therefore

$\frac{1}{2}m{k}^{2}2gR=\frac{mg(r-R)}{1+\frac{(r-R)}{R}}$

$k^{2}R[1+\frac{r-R}{R}]=r-R$

or $k^{2}r=r-R$

or $r=\frac{R}{1-k^2}$